Explanation

(a)
Oxalic acid = $x$ mol/L
Oxalic acid $KMnO_{4}$
$M_{1}V_{1} = M_{2}V_{2}$
$40\ mL \times x = 16\ mL \times 0.05$
$x = \frac{16 \times 0.05}{40} = \frac{1}{50}$
$x = \frac{1}{50}$ M
Now, convert molarity into normality
$N \times eq.wt = M \times mol.wt.of\ oxalic\ acid$
$N \times 45 = \frac{1}{50} \times 90$
$N = \frac{1}{25}$
This normality represents the hydrogen ion concentration.
So, $\left\lbrack H^{+} \right\rbrack = \frac{1}{25}$
$pH = \log\frac{1}{\lbrack H^{+}\rbrack}$
$= \log{25 = 1.3}$